= Solution
For the simple-contour form of the <residue theorem>, let $C$ be a positively oriented, simple, closed, piecewise smooth <contour>. Assume $f$ is <holomorphic> on a neighbourhood of $C$ and its interior except at finitely many isolated singularities $z_1,\ldots,z_s$, all strictly inside $C$, with none on $C$. Then
$$
\oint_C f(z)\,dz=2\pi i\sum_{j=1}^s\operatorname{Res}_{z=z_j}f.
$$
The more general version weights each <residue> by the contour's <winding number>, with the usual requirement that the contour be homologous to zero in the domain of analyticity containing it and the nonsingular region.
For the requested application, assume $f$ is <holomorphic> on a neighbourhood of the contour and its closed interior, and $z_0$ is strictly inside it. Its <Taylor series> shows
$$
\frac{f(z)}{(z-z_0)^2}
=\frac{f(z_0)}{(z-z_0)^2}+\frac{f'(z_0)}{z-z_0}+O(1).
$$
Only the second term contributes a <residue>. The <residue theorem> therefore proves the derivative form of the <Cauchy integral formula>:
$$
\boxed{\frac1{2\pi i}\oint_C\frac{f(z)}{(z-z_0)^2}\,dz=f'(z_0)}.
$$
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